`int 1/sqrt(16-x^2) dx` Find the indefinite integral

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Given to solve ,

`int 1/sqrt(16-x^2) dx`

using the Trig substitutions

for `sqrt(a-bx^2)`

`x= sqrt(a/b) sin(u)`

so for ,

`int 1/sqrt(16-x^2) dx` --------(1)

so , `x` can be

`x= sqrt(16/1) sin(u)`

 = `4sin(u)`

=> `dx = 4cos(u) du`

so,for (1) we get

`int 1/sqrt(16-(4sin(u))^2) (4cos(u) du)`

=`int (4cos(u))/sqrt(16-16(sin(u))^2) du`

=`...

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Given to solve ,

`int 1/sqrt(16-x^2) dx`

using the Trig substitutions

for `sqrt(a-bx^2)`

`x= sqrt(a/b) sin(u)`

so for ,

`int 1/sqrt(16-x^2) dx` --------(1)

so , `x` can be

`x= sqrt(16/1) sin(u)`

 = `4sin(u)`

=> `dx = 4cos(u) du`

so,for (1) we get

`int 1/sqrt(16-(4sin(u))^2) (4cos(u) du)`

=`int (4cos(u))/sqrt(16-16(sin(u))^2) du`

=` int (4cos(u))/(4sqrt(1-sin^2(u))) du`

= `int (4cos(u))/(4sqrt(cos^2(u))) du`

=  `int (4cos(u))/(4cos(u)) du`

=` int (1) du`

=` u+c`

but `x= 4sin(u)`

=> `x/4 = sin(u)`

=> `u = arcsin(x/4)`

so ,

=> `u+c`

= `arcsin(x/4)+c`

so ,

`int 1/sqrt(16-x^2) dx = arcsin(x/4)+c`

 

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